Chemistry 9701/35 — October/November 2015

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment you will determine the ionic equation for the reaction of acidified potassium manganate(VII) with potassium iodide. Excess potassium iodide is used and the reaction produces iodine. The amount of iodine produced is measured by titration with sodium thiosulfate.

FA 1 is $0.0180\text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FA 2 is $1.00\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .
FA 3 is $0.500\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FA 4 is $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
starch indicator

(a)

Method

Pipette $25.0\text{ cm}^3$ of FA 1 into a conical flask.
Use the measuring cylinder to add $25\text{ cm}^3$ of FA 2 to the conical flask.
Use the measuring cylinder to add $20\text{ cm}^3$ of FA 3 to the conical flask.
Fill the burette with FA 4.
Carry out a rough titration. When the colour of the mixture becomes yellow/orange, add a few drops of starch indicator. Then titrate until the mixture goes colourless.
Record all your burette readings in the space below.

The rough titre is ........................ $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record in a suitable form below all of your burette readings and the volume of FA 4 added in each accurate titration.

Keep FA 1 and FA 2 for use in Question 3 and FA 4 for use in Question 2.

DifficultyMedium-Easy

Worked solution

Answer

Record a rough titre and at least two accurate titrations. All burette readings should be to the nearest $0.05\text{ cm}^3$ .

Example record (candidate-dependent; your own readings will be used):

rough titre = $22.90\text{ cm}^3$

Titration	initial burette reading / $\text{cm}^3$	final burette reading / $\text{cm}^3$	titre / $\text{cm}^3$
1	0.00	22.50	22.50
2	22.50	45.00	22.50

Final answer

Candidate-dependent; example accurate titres 22.50 cm3 and 22.50 cm3

Detailed explanation

Background Concept

This experiment is an iodine/thiosulfate redox titration. Acidified potassium manganate(VII) oxidises excess potassium iodide to iodine, and the iodine produced is then titrated with sodium thiosulfate. Starch forms an intense blue-black complex with iodine, so a few drops are added near the endpoint; the endpoint is reached when the blue-black colour disappears.

Understanding the Question

Part (a) is a practical recording task. It does not ask for a calculation. The marks are awarded for carrying out the titration properly and recording the burette data clearly: at least two accurate titrations, readings to the nearest 0.05 cm3, concordant titres within 0.10 cm3, and a table with headings and units.

Approach

First do a rough titration to find the approximate endpoint. Then repeat the titration carefully, adding starch only when the solution becomes yellow/orange, and continue until the solution is colourless. Record initial and final burette readings and calculate each titre as final minus initial. Repeat until two accurate titres agree closely.

Step-by-Step Reasoning

Pipette 25.0 cm3 of FA1 into the conical flask.
Use a measuring cylinder to add 25 cm3 of FA2 and 20 cm3 of FA3.
Fill the burette with FA4 and do a rough titration, recording the rough titre.
For each accurate titration, run in FA4 until the colour is yellow/orange, add a few drops of starch, then titrate until colourless.
Record all readings to the nearest 0.05 cm3 and calculate titre = final reading - initial reading.
Repeat until two accurate titres are within 0.10 cm3 of each other.

The example table shows one possible set of results: both accurate titres are 22.50 cm3.

Key Takeaways

Correct practical recording requires units, headings, appropriate precision, and enough repeated measurements to judge reliability. Concordant titres are the basis of a reliable titration result.

Common Mistakes

Writing table headings without units.
Writing 'amount' instead of 'volume' for burette readings.
Recording burette readings to only 1 decimal place or to 0.01 cm3.
Performing only one accurate titration.
Using 50.00 cm3 as an initial burette reading.
Adding starch at the very start of the titration instead of near the endpoint.

Things to Be Careful About

All burette readings should be to the nearest 0.05 cm3. The titre is always final minus initial reading. At least two accurate titrations are needed, and they should be concordant. Keep FA1, FA2 and FA4 for later questions.

Techniques used

perform a rough titrationrecord burette readings to the nearest 0.05 cm3add starch indicator near the endpointrepeat accurate titrations until concordanttabulate data with headings and units

(b)

From your accurate titration results, obtain a suitable value for the volume of FA 4 to be used in your calculations.
Show clearly how you have obtained this value.

Volume of FA 4 required is ..................... $\text{cm}^3$ .

DifficultyEasy

Worked solution

Working

Select two concordant accurate titres within a total spread of $0.20\text{ cm}^3$ .

Example: $22.50\text{ cm}^3$ and $22.50\text{ cm}^3$ .

\text{mean} = \frac{22.50 + 22.50}{2} = 22.50\text{ cm}^3

Answer

Volume of FA 4 = $22.50\text{ cm}^3$ (candidate-dependent example).

Final answer

22.50 cm3 (candidate-dependent example)

Detailed explanation

Background Concept

The mean titre is the value used in all subsequent calculations. It should be based only on accurate, concordant titres, not on the rough titre or on poorly agreeing results.

Understanding the Question

Part (b) asks you to choose your best accurate results and show clearly how you obtained a single working volume of FA4. The mark is for selecting appropriate titres and calculating a mean correctly.

Approach

Identify two (or more) accurate titres that are close together. Their total spread should be no more than 0.20 cm3. Average them and quote the mean to the nearest 0.01 cm3, unless an allowed special case applies.

Step-by-Step Reasoning

Look at the accurate titres recorded in part (a).
Select two or more that are concordant, e.g. 22.50 and 22.50 cm3.
Add them and divide by the number of values.
Here the mean is exactly 22.50 cm3, already correct to 2 decimal places.
Use this mean as the volume of FA4 in the calculations.

Key Takeaways

A reliable mean titre is based only on concordant results, and its precision should match sensible titration precision.

Common Mistakes

Including the rough titre in the mean.
Averaging results that are more than 0.20 cm3 apart.
Quoting the mean to 3 decimal places when it is not an allowed special case.
Not showing which titres were selected.

Things to Be Careful About

The mean should normally be quoted to 2 decimal places. If two identical titres such as 22.50 and 22.50 are used, the mean is exactly 22.50 cm3.

Techniques used

select concordant accurate titrescalculate the mean titreround the mean to an appropriate precision

(c)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Calculate the number of moles of sodium thiosulfate in the volume of FA 4 calculated in (b).

moles of $\text{Na}_2\text{S}_2\text{O}_3$ = ............................. mol

DifficultyEasy

Worked solution

Working

$n(\text{Na}_2\text{S}_2\text{O}_3) = 0.100 \times \frac{22.50}{1000} = 0.00225\text{ mol}$

Answer

$0.00225$ mol

Final answer

0.00225 mol

Detailed explanation

Background Concept

The amount of solute in moles is found from concentration and volume:

$n = c \times V$

where the volume must be in dm3.

Understanding the Question

This part uses the mean titre from part (b) to calculate the number of moles of sodium thiosulfate used in the titration.

Approach

Convert the volume of FA4 from cm3 to dm3 by dividing by 1000, then multiply by the concentration of FA4.

Step-by-Step Reasoning

Using the example volume of 22.50 cm3:

$n(\text{Na}_2\text{S}_2\text{O}_3) = 0.100 \times \frac{22.50}{1000}$

$= 0.00225\text{ mol}$

Key Takeaways

Volume in cm3 must be divided by 1000 before using the formula n = cV. The concentration is in mol dm-3.

Common Mistakes

Forgetting to convert cm3 to dm3.
Quoting too many significant figures.
Using the wrong concentration.

Things to Be Careful About

The final answer should be given to 3 or 4 significant figures. Here 0.00225 mol has 3 significant figures.

Techniques used

convert volume from cm3 to dm3calculate moles from concentration and volume

(ii)

Use the equation below to calculate the number of moles of iodine that reacted with the sodium thiosulfate in the titration.

\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}

moles of $\text{I}_2$ = ............................. mol

DifficultyEasy

Worked solution

Working

From the equation, $1\text{ mol I}_2$ reacts with $2\text{ mol Na}_2\text{S}_2\text{O}_3$ .

n(\text{I}_2) = \frac{0.00225}{2} = 0.001125\text{ mol}

Answer

$0.001125$ mol

Final answer

0.001125 mol

Detailed explanation

Background Concept

The equation shows that one mole of iodine reacts with two moles of sodium thiosulfate:

$\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}$

So the mole ratio of I2 to Na2S2O3 is 1:2.

Understanding the Question

Part (ii) asks you to use the moles of thiosulfate from part (i) to calculate the moles of iodine titrated.

Approach

Divide the moles of thiosulfate by 2 because the ratio is 1 I2 : 2 thiosulfate.

Step-by-Step Reasoning

Using 0.00225 mol of thiosulfate:

$n(\text{I}_2) = \frac{0.00225}{2} = 0.001125\text{ mol}$

Key Takeaways

Stoichiometric calculations require reading the mole ratio directly from the balanced equation.

Common Mistakes

Multiplying by 2 instead of dividing by 2.
Using the wrong equation or ratio.
Rounding too early and losing accuracy.

Things to Be Careful About

The final answer 0.001125 mol has 4 significant figures, which is acceptable. Keep the unrounded value for later parts where possible.

Techniques used

apply the stoichiometric ratio I2:thiosulfate = 1:2calculate moles of iodine

(iii)

Use information on page 2 to calculate the number of moles of potassium manganate(VII) in FA 1 used in the titration.

moles of $\text{KMnO}_4$ = ............................. mol

DifficultyEasy

Worked solution

Working

$n(\text{KMnO}_4) = 0.0180 \times \frac{25.0}{1000} = 0.000450\text{ mol}$

Answer

$0.000450$ mol

Final answer

0.000450 mol

Detailed explanation

Background Concept

The pipette delivers 25.0 cm3 of FA1, which is 0.0250 dm3. The concentration of FA1 is 0.0180 mol dm-3.

Understanding the Question

Part (iii) asks for the number of moles of potassium manganate(VII) in the FA1 used in the titration flask.

Approach

Use n = cV, with the volume converted to dm3.

Step-by-Step Reasoning

$n(\text{KMnO}_4) = 0.0180 \times \frac{25.0}{1000} = 0.000450\text{ mol}$

Key Takeaways

The pipette volume is quoted to one decimal place, so the calculation should reflect the given precision.

Common Mistakes

Using 25.0 cm3 as if it were dm3.
Mixing up FA1 and FA4 concentrations.
Quoting too many significant figures.

Things to Be Careful About

0.000450 mol has three significant figures. The trailing zero after 45 is significant because it follows a decimal point.

Techniques used

calculate moles of KMnO4 from concentration and pipetted volume

(iv)

From your answers to (ii) and (iii), calculate the number of moles of iodine produced by the reaction of $2.00$ moles of potassium manganate(VII) with excess potassium iodide.

moles $\text{I}_2$ = ............................. mol

DifficultyMedium-Easy

Worked solution

Working

$n(\text{I}_2\text{ for }2.00\text{ mol KMnO}_4) = \frac{0.001125}{0.000450} \times 2.00 = 5.0\text{ mol}$

Answer

$5.0$ mol

Final answer

5.0 mol

Detailed explanation

Background Concept

The result from the titration tells us how many moles of iodine are produced by 0.000450 mol of KMnO4. To find how much iodine 2.00 mol of KMnO4 would produce, scale up by the ratio of moles of KMnO4.

Understanding the Question

Part (iv) asks you to use your answers to (ii) and (iii) to calculate the moles of iodine produced by 2.00 mol of KMnO4. This ratio will identify the correct ionic equation in part (v).

Approach

Divide moles of I2 by moles of KMnO4 to find the moles of I2 per mole of KMnO4, then multiply by 2.00.

Step-by-Step Reasoning

Using the example values:

$\frac{0.001125}{0.000450} = 2.5$

So 1.00 mol of KMnO4 produces 2.5 mol of I2. Therefore 2.00 mol of KMnO4 produces:

$2.5 \times 2.00 = 5.0\text{ mol I}_2$

Key Takeaways

A measured ratio can be scaled to any target amount of reactant. The theoretical value here is 5.0 mol I2 per 2 mol KMnO4.

Common Mistakes

Dividing the wrong way round.
Forgetting to multiply by 2.00.
Rounding intermediate values too aggressively.

Things to Be Careful About

The final answer should be 5.0 mol, consistent with the expected stoichiometry of the reaction.

Techniques used

scale the moles of iodine to 2.00 mol KMnO4interpret a stoichiometric ratio

(v)

Using your answer to (iv), put a tick next to the ionic equation that represents the reaction between FA 1 and FA 3.

$2\text{MnO}_4^- + 2\text{I}^- + 16\text{H}^+ \rightarrow \text{I}_2 + 2\text{Mn}^{6+} + 8\text{H}_2\text{O}$ [ ]
$2\text{MnO}_4^- + 4\text{I}^- + 16\text{H}^+ \rightarrow 2\text{I}_2 + 2\text{Mn}^{5+} + 8\text{H}_2\text{O}$ [ ]
$2\text{MnO}_4^- + 6\text{I}^- + 16\text{H}^+ \rightarrow 3\text{I}_2 + 2\text{Mn}^{4+} + 8\text{H}_2\text{O}$ [ ]
$2\text{MnO}_4^- + 8\text{I}^- + 16\text{H}^+ \rightarrow 4\text{I}_2 + 2\text{Mn}^{3+} + 8\text{H}_2\text{O}$ [ ]
$2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 5\text{I}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$ [ ]
$2\text{MnO}_4^- + 12\text{I}^- + 16\text{H}^+ \rightarrow 6\text{I}_2 + 2\text{Mn}^+ + 8\text{H}_2\text{O}$ [ ]

DifficultyMedium-Easy

Worked solution

Answer

Tick the equation:

$2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 5\text{I}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

Final answer

2MnO4^- + 10I^- + 16H^+ -> 5I2 + 2Mn2+ + 8H2O

Detailed explanation

Background Concept

The correct ionic equation must show the real redox chemistry: MnO4- is reduced to Mn2+ in acid, while I- is oxidised to I2. Since the calculated result in part (iv) is 5.0 mol I2 per 2.00 mol KMnO4, the equation must have 5I2 on the right.

Understanding the Question

Part (v) asks you to select the equation consistent with the moles of iodine calculated in part (iv). The equations differ in the number of I- ions used, the number of I2 molecules produced, and the oxidation state of manganese in the product.

Approach

Use the result from part (iv) to choose the equation with 5I2 per 2MnO4-. Then check that the equation is balanced and that the manganese product is Mn2+, the normal product of acidified MnO4-.

Step-by-Step Reasoning

The half-equations are:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

$2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$

To balance electrons, multiply the first by 2 and the second by 5:

$2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

$10\text{I}^- \rightarrow 5\text{I}_2 + 10\text{e}^-$

Adding these gives:

$2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 5\text{I}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

This matches the fifth option and the 5.0 mol result from part (iv).

Key Takeaways

The correct equation must be stoichiometrically balanced and chemically plausible. MnO4- in acid is reduced to Mn2+, not Mn6+, Mn5+, Mn4+, Mn3+ or Mn+.

Common Mistakes

Choosing an equation based only on balancing atoms but with an impossible manganese oxidation state.
Ignoring the calculated stoichiometry from part (iv).
Forgetting that H+ and H2O must balance too.

Things to Be Careful About

The number of I- ions must provide the electrons needed to reduce MnO4- to Mn2+. For 2 MnO4-, 10 electrons are needed, supplied by 10 I- ions, producing 5 I2.

Techniques used

compare calculated moles of I2 per 2 mol KMnO4 with the equationsidentify the correct redox stoichiometry

(vi)

Prove that the iodide ion has been oxidised in the equation that you selected in (v).

DifficultyMedium-Easy

Worked solution

Answer

Each iodide ion loses one electron:

$2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$

The oxidation number of iodine increases from $-1$ in $\text{I}^-$ to $0$ in $\text{I}_2$ , so iodide is oxidised.

Final answer

Oxidation number of I increases from -1 to 0; 2I- -> I2 + 2e-

Detailed explanation

Background Concept

Oxidation is loss of electrons. In terms of oxidation number, oxidation is an increase in oxidation number.

Understanding the Question

Part (vi) asks you to prove that the iodide ion has been oxidised in the equation you selected. You need to state either the half-equation or the oxidation number change.

Approach

Look at iodine before and after the reaction. In I- it has oxidation number -1; in I2 it has oxidation number 0. Since the oxidation number increases, oxidation has occurred. Equivalently, iodide ions lose electrons.

Step-by-Step Reasoning

In I-, each iodine atom has oxidation number -1.
In I2, each iodine atom has oxidation number 0.
The change is from -1 to 0, an increase of 1 per iodine atom.
Loss of electrons is shown by the half-equation:

$2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$

This satisfies the definition of oxidation.

Key Takeaways

Oxidation number increase and electron loss are two equivalent ways of identifying oxidation. The balanced half-equation is a precise way to prove it.

Common Mistakes

Saying iodine gains electrons.
Giving an unbalanced half-equation.
Confusing oxidation number decrease with oxidation.

Things to Be Careful About

The half-equation must be balanced in both atoms and charge: two I- on the left provide two negative charges, and I2 plus two electrons on the right also gives total charge -2.

Techniques used

deduce the oxidation number change of iodineidentify loss of electrons in oxidation

(d)

(i)

The error in calibration of the pipette you used is $\pm 0.06\text{ cm}^3$ .
Calculate the percentage error when measuring FA 1, using the pipette.

percentage error = ..................... %

DifficultyEasy

Worked solution

Working

$\text{percentage error} = \frac{0.06}{25.0} \times 100 = 0.24\%$

Answer

$0.24\%$

Final answer

0.24%

Detailed explanation

Background Concept

Percentage error expresses the uncertainty of a measurement as a percentage of the measured value:

$\text{percentage error} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100$

Understanding the Question

The pipette used to measure 25.0 cm3 of FA1 has a calibration error of ±0.06 cm3. You must calculate the percentage error for this single volume measurement.

Approach

Use the given absolute error and the pipette volume as the measured value.

Step-by-Step Reasoning

$\text{percentage error} = \frac{0.06}{25.0} \times 100 = 0.24\%$

Key Takeaways

Percentage error is useful for comparing the precision of different measurements.

Common Mistakes

Dividing 0.06 by 1000 instead of by 25.0.
Forgetting to multiply by 100.
Giving only 0.24 without the percentage sign.

Things to Be Careful About

The measured value is 25.0 cm3, not 25.0 dm3. The answer is 0.24%.

Techniques used

calculate percentage erroruse absolute uncertainty and measured value

(ii)

A student suggested that the experiment would be more accurate if a pipette was used to measure solution FA 3.
State and explain whether you agree with the student.

DifficultyMedium-Easy

Worked solution

Answer

Disagree. FA 3 / potassium iodide is present in excess, so a small variation in its volume does not affect the amount of iodine produced. The amount of iodine is limited by FA 1, so the measuring cylinder is acceptable for FA 3.

Final answer

Disagree: KI is in excess

Detailed explanation

Background Concept

In the reaction, KMnO4 is the limiting reagent and KI is present in excess. As long as enough iodide is present to react with all the KMnO4, the amount of iodine produced depends on the amount of KMnO4, not on the exact amount of KI.

Understanding the Question

The student suggests using a pipette to measure FA3 to improve accuracy. You must decide whether the suggestion is valid.

Approach

Consider whether the volume of FA3 affects the quantity being measured. Since KI is in excess, its exact volume is not critical.

Step-by-Step Reasoning

The amount of iodine produced is determined by the amount of KMnO4 in FA1.
FA3 is added in excess to ensure all KMnO4 reacts.
A small error in the volume of FA3 therefore does not change the moles of iodine produced.
Using a pipette would improve precision of a measurement that does not affect the result, so the suggestion is not necessary.

Key Takeaways

Marking an improvement as useful requires checking whether the quantity being measured actually affects the final result.

Common Mistakes

Agreeing automatically with any suggestion that uses a more precise apparatus.
Not mentioning that KI is in excess.
Saying the volume of FA3 affects the moles of iodine.

Things to Be Careful About

The correct answer is that the student is wrong because FA3/KI is in excess.

Techniques used

evaluate the effect of excess reagent on precisionidentify the limiting reagent

The rest of this paper

2 more questions

Q2Presentation of Data and Observations · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation9M
Q3Qualitative Analysis · Manipulation, Measurement and Observation9M

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